1. $$ P(X)=\frac{1}{5} $$, for $$ X=0,1,2,3,4,5 $$

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It seems like you are providing a probability distribution function $$ P(X) $$ for a discrete random variable $$ X $$ that takes on values from 0 to 5. The function is defined as $$ P(X) = \frac{1}{5} $$ for each of these values.

Let's analyze this step by step:

1. **Identify the values of $$ X $$**: The random variable $$ X $$ can take the values $$ 0, 1, 2, 3, 4, 5 $$.

2. **Probability Distribution**: The probability for each value of $$ X $$ is given as $$ P(X) = \frac{1}{5} $$. This means that each outcome has an equal probability of occurring.

3. **Check the total probability**: The sum of the probabilities for all possible values of $$ X $$ should equal 1. We can calculate this as follows:

$$
P(0) + P(1) + P(2) + P(3) + P(4) + P(5) = \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{5}
$$

Calculating this gives:

$$
6 \times \frac{1}{5} = \frac{6}{5}
$$

Since $$ \frac{6}{5} $$ is greater than 1, this indicates that the probabilities do not sum to 1, which violates the fundamental property of probability distributions.

### Conclusion
The provided probability distribution $$ P(X) = \frac{1}{5} $$ for $$ X = 0, 1, 2, 3, 4, 5 $$ is incorrect because the total probability exceeds 1.

To correct this, the probabilities should be adjusted so that they sum to 1. Since there are 6 outcomes, each probability should be:

$$
P(X) = \frac{1}{6}
$$

for $$ X = 0, 1, 2, 3, 4, 5 $$.

If you have any further questions or need additional calculations, feel free to ask!
The probability distribution $$ P(X) = \frac{1}{5} $$ for $$ X = 0, 1, 2, 3, 4, 5 $$ is incorrect because the total probability exceeds 1. Each probability should be $$ \frac{1}{6} $$ to sum to 1.

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